A car traveling at 50 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 74 cm (with respect to the road) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of 43 kg?

First find the acceleration from

u = 50km/h = 50*1000/3600 m/s = 13.89 m/s
to
v = 0 m/s
through a distance of
S = 74 cm = 0.74 m

v²-u^sup2;=2aS
a = (v²-u²)/2S
=-13.89²/(2*0.74)
= 130.34 m/s² (approx. 13g)

F=ma
=43 * 130.34 N
= 5604.6 N

To find the magnitude of force acting on the passenger, we can use the impulse-momentum principle.

Step 1: Convert the velocity of the car from km/h to m/s.
Given: Velocity of the car = 50 km/h
1 kilometer = 1000 meters
1 hour = 3600 seconds

50 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 13.89 m/s

Step 2: Convert the distance moved by the passenger from centimeters to meters.
Given: Distance moved by the passenger = 74 cm
1 meter = 100 centimeters

74 cm * (1 m / 100 cm) = 0.74 m

Step 3: Calculate the initial momentum of the passenger.
The initial momentum of the passenger can be calculated using the formula:

Momentum = mass * velocity

Given: Mass of the passenger = 43 kg
Velocity of the car = 13.89 m/s

Initial momentum = 43 kg * 13.89 m/s = 597.27 kg·m/s

Step 4: Calculate the final momentum of the passenger.
The final momentum of the passenger can be calculated using the formula:

Final momentum = 0 (as the passenger comes to rest)

Step 5: Calculate the change in momentum.
Change in momentum = Final momentum - Initial momentum
Change in momentum = 0 - 597.27 kg·m/s = -597.27 kg·m/s

Step 6: Calculate the force exerted on the passenger.
Force = Change in momentum / time

Since the force acting on the passenger is assumed constant, the time taken for the passenger to come to rest can be determined by the equation:

Distance = (Initial velocity * time) + (0.5 * acceleration * time^2)

Rearranging this equation, we can solve for time:

0.74 m = (13.89 m/s * time) + (0.5 * acceleration * time^2)

We know that acceleration is the change in velocity divided by time:

Acceleration = Change in velocity / time

In this case, the change in velocity is 13.89 m/s (from the car's initial velocity to zero) and the time is the unknown we are trying to solve for. Therefore:

13.89 m/s = (13.89 m/s - 0 m/s) / time

Simplifying the equation, we find:

time = (13.89 m/s) / (13.89 m/s) = 1 second

Substituting the time into the original equation:

0.74 m = (13.89 m/s * 1 s) + (0.5 * acceleration * (1 s)^2)

0.74 m = 13.89 m/s + 0.5 * acceleration

Rearranging the equation to solve for acceleration:

0.37 m = 0.5 * acceleration

acceleration = (2 * 0.37 m) / 1 s^2 = 0.74 m/s^2

Now, substituting the values into the equation for force:

Force = (-597.27 kg·m/s) / 1 s = -597.27 N

Since impulse and force are vectors, the negative sign indicates that the force acts in the opposite direction (opposite to the initial momentum of the passenger).

Therefore, the magnitude of the force acting on the passenger's upper torso is 597.27 N.

To determine the magnitude of the force acting on the passenger's upper torso, we can use the concept of deceleration.

First, we need to determine the initial velocity of the passenger. Since the car hits the bridge abutment, we can assume that both the car and the passenger were initially moving at the same velocity, which is 50 km/h.

Converting the initial velocity of the passenger from km/h to m/s:
50 km/h = 50,000 m/3600 s = 13.9 m/s

Next, we need to determine the final velocity of the passenger. The passenger is brought to rest by an inflated airbag, so the final velocity is 0 m/s.

Using the equation of motion: v^2 = u^2 + 2as
where v is the final velocity (0 m/s), u is the initial velocity (13.9 m/s), a is the deceleration, and s is the distance travelled (74 cm = 0.74 m).

Rearranging the equation, we have:
0 = (13.9 m/s)^2 + 2a * 0.74 m

Simplifying and solving for deceleration (a):
a = -[(13.9 m/s)^2] / [2 * 0.74 m]
a = -127.81 m/s^2

The negative sign indicates that the acceleration is in the opposite direction of motion, representing deceleration.

Lastly, we can use Newton's second law of motion (F = ma) to find the magnitude of the force acting on the passenger's upper torso.

F = m * a
F = 43 kg * (-127.81 m/s^2)
F ≈ -5508.83 N

The magnitude of the force acting on the passenger's upper torso is approximately 5508.83 N. The negative sign indicates that the force is directed opposite to the motion of the passenger.